Question

1. $lim_{s

ightarrow - 1}\frac{s^{2}+6s + 5}{s^{2}-3s - 4}$ * hint: change the s’s into x’s and do the x game for both the top and bottom

Explanation:

Step1: Factor the numerator

Factor $s^{2}+6s + 5$ as $(s + 1)(s+5)$.

Step2: Factor the denominator

Factor $s^{2}-3s - 4$ as $(s + 1)(s - 4)$.

Step3: Simplify the fraction

The original limit $\lim_{s
ightarrow - 1}\frac{s^{2}+6s + 5}{s^{2}-3s - 4}=\lim_{s
ightarrow - 1}\frac{(s + 1)(s + 5)}{(s + 1)(s - 4)}$. Cancel out the common - factor $(s + 1)$ (since $s
eq - 1$ when taking the limit), we get $\lim_{s
ightarrow - 1}\frac{s + 5}{s - 4}$.

Step4: Substitute the value of s

Substitute $s=-1$ into $\frac{s + 5}{s - 4}$, we have $\frac{-1 + 5}{-1-4}=\frac{4}{-5}=-\frac{4}{5}$.

Answer:

$-\frac{4}{5}$